Quant Boosters - Vikas Saini - Set 1



  • let 6th month he had a sale of z.
    6435+6927+6855+7230+6562+z = 6 x 6500
    = > 34009 + z = 39000
    = > z = 4991.

    Short approach :-
    let's take 6500 as base.
    6500-(-65+427+355+730+62)
    =6500 - (1509)
    = 4991.



  • Q6) The average weight of boys in a school is 47 and girls is 42. If there are 6 more boys than girls and average weight of all class is 44.8. How many students are in the class?



  • let no of girls = a
    no of boys = a+6
    47(a+6)+42a = 44.8(a+a+6)
    47a+42a+282 = 44.8(2a+6)
    89a + 282 = 89.6a + 268.8
    0.6 a = 13.2
    a = 22.
    a+6 = 28.
    total = 22+28 = 50.

    Short approach :-
    average of all class=44.8.
    ratio of boys & girls= (47-44.8 ):(44.8-42)=2.2:2.8.
    (2.8-2.2)x=6
    x=10.
    total students= (2.2+2.8 )x10=50



  • Q7) Average weight of 11 players is increased by 1kg. When one player of team weighing 55kg is replaced by new player. Find weight of new player



  • Average weight of each player = a.
    weight of new player = k.
    11a - 55 + k = 11(a+1)
    k - 55 = 11
    k = 66.

    Short approach :-
    weight of new player = 55 + 1 x 11 = 66.



  • Q8) The average of 71 results is 48. If the average of first 59 results is 46 and that of the last 11 is 52 then find the 60th result.



  • 71 x 48 = 59 x 46 + 11 x 52 + 60th
    = > 60th value = 71 x 48 - 59 x 46 - 11 x 52.
    = > 60th value = 122.

    Short approach :-
    We take 48 as base.
    59 results average is -2 (from base)
    11 results average is +4 (from base)
    So 59 x (-2) + 11 x 4 = -74.
    To get 60th value this value should be decrease from base.
    Hence 60th value = 48 - (-74) = 122.



  • Q9) The average of 50 numbers is 38. If two numbers, namely 45 & 55 are discarded then what is the average of remaining numbers ?



  • Total = 50 x 38 = 1900.
    two numbers discarded (45,55)
    new total = 1900-45-55 = 1800.
    new average = 1800 / 48 = 37.5

    Short approach :-
    The average of 50 numbers is 38.
    so if we discarded two number then to maintain average of 38,value should be decreased by 2 x 38 = 76.
    but here value decreased by 100.
    So 100-76 = 24.
    There is 48 numbers.
    so every number value will be decreased by 24/48 = 0.5
    new average = 38 - 0.5 = 37.5.



  • Q10) The average runs of Rohit & Dhawan is 280. The average runs of Virat & Rahane is 200. What is the average of each of them.



  • Rohit + Dhawan = 2 x 280 = 560.
    Virat + Rahane = 2 x 200 = 400.
    Total runs = 560 + 400 = 960.
    Average = 960 / 4 = 240.

    Short approach :-
    Take average of any two
    let we take Rohit & Dhawan,which is nothing but 280.(base)
    now average of virat & rahane 200.
    which is -80 (from base)
    Average = 280 - (80/2)
    = 280 - 40
    = 240.

    We can solve it by take 200 too..
    base = 200.
    average of Rohit & Dhawan is +80(from base).
    average = 200 + (80/2)
    = 240.
    Here divided by 2, because that is average of two person..



  • Q11) The average weight of a class of 29 students is 40 kg. If the weight of teacher be included,the average rises by 500gm. What is the weight of the teacher ?



  • previous weight of class = 29 x 40 = 1160 kg.
    new weight (including teacher) = 30 x 40.5 = 1215.
    Teacher's weight = 1215 - 1160 = 55 kg.

    Short approach :-
    base = 40.
    including teacher average increased by 0.5 kg.
    Teacher's weight = 40 + 30 x 0.5
    = 40 + 15
    = 55 kg.



  • Q12) By selling 18 oranges profit is equal to SP of 3 oranges. Find profit percentage.



  • Sp of 1 orange = x.
    cp of 1 orange = y.
    18x – 18y = 3x
    = > 18 y = 15 x.
    Profit percentage = (3x/18y)x100
    = (3x/15x)x100
    = 20%.

    Short approach :-
    Profit percentage = [profit (quantity) / (Total quantity – profit)] x 100
    = [(3)/(18-3)] x 100
    = [3/15] x 100
    = 20%



  • Q13) By selling 18 chocolates, vendor loss selling price of 2 chocolates. Find loss percentage.



  • sp of 1 chocolate = x
    cp of 1 chocolate = y
    18 y – 18 x = 2x.
    18 y = 20 x.
    Loss percentage = (2x/18 y) x 100
    = (2x/20x)x100
    = 10%.

    Short approach :-
    Loss percentage = [Loss (quantity) / (Total quantity + loss) ] x 100
    = [ 2 / (18+2) ] x 100
    = 10%.



  • Q14) On selling an article for rs 576 a trader losses 4%. In order to gain 25/6 % , he must sell article for
    a. 655 rs
    b. 676 rs
    c. 625 rs
    d. 600 rs.



  • Cost price = [576 / (100 – 4)] x 100 = 600 rs.
    Selling price = 600 x (1 + 25/(6 x 100) = 625.

    Short approach :-
    Direct formula to find sp in such problems
    = previous selling price x 100 x (1+ gain percent)/(1 – loss percent)
    = 576 x 100 x (1+25/600)/(1 – 4/100)
    = 625 rs.



  • Q15) A goldsmith has 361 rings of gold. He sells some of them at a loss of 4% and rest at a profit of 15%. Overall profit is 8%. Find no of rings sold at profit of 15%.


Log in to reply